Asked by Anonymous
If an object is initially moving with a constant velocity of 20m/s towards positive x-axis and after 5 sec it changes its direction and moving at an angle of 30o with x-axis, find magnitude and direction of its acceleration.
Answers
Answered by
Henry
Vo = 20 m/s[0o]
V = 20 m/s[30o]
a = (V-Vo)/t = (20[30o]-20[0o])/5 =
(20*cos30+i20*sin30-20)/5 =
(17.32+10i-20)/5 = (-2.68+10i)/5 =
-0.536 + 2i,(Q2).
Tan Ar = Y/X = 2/-0.536 = -3.73134
Ar = -75o = Reference angle.
A = -75 + 180 = 105o,CCW = 15o W. of N.
a = 2/sin105 = 2.07 m/s^2 @ 105o, CCW.
V = 20 m/s[30o]
a = (V-Vo)/t = (20[30o]-20[0o])/5 =
(20*cos30+i20*sin30-20)/5 =
(17.32+10i-20)/5 = (-2.68+10i)/5 =
-0.536 + 2i,(Q2).
Tan Ar = Y/X = 2/-0.536 = -3.73134
Ar = -75o = Reference angle.
A = -75 + 180 = 105o,CCW = 15o W. of N.
a = 2/sin105 = 2.07 m/s^2 @ 105o, CCW.
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